Chain of Faith: Policy-Internal Rewards for Reasoning Reliance
Organizations: University of California, Davis
Abstract
Chain-of-thought (CoT) explanations can sound plausible while having little influence on a model's final answer. We study four rewards that use the policy's own signals to encourage reliance on generated reasoning: gradient sensitivity, confidence gaps, progressive confidence gains, and reasoning--prompt swaps. We add each reward to group relative policy optimization (GRPO) and evaluate two small instruction-tuned models on GSM8K and StrategyQA using reasoning edits, paraphrases, and prompt hints. Several configurations increase answer reliance to reasoning edits, while effects on accuracy and paraphrase robustness vary. We also observe answer copying at a high reward weight and high reliance in decomposition settings with poor task accuracy. The study compares practical objectives for reasoning reliance, confirms the trade-off between reliance and performance, and proposes four lightweight rewards that can improve reasoning reliance.
Figures & tables
| Reward | Signal and geometry | Intervention context | Additional computation | Principal limitation |
|---|---|---|---|---|
| Mean token gradient norm ( ) | Local continuous embedding changes | Answer-likelihood backward pass | Surface sensitivity and answer copying | |
| Signed likelihood gap / Frobenius distance | Reasoning removed | One answer-likelihood forward evaluation | Context changes and answer leakage | |
| Positive prefix gains / path length | Answer forced at step boundaries | Answer evaluation at cached prefixes | Confidence oscillations and prefix context changes | |
| Ratio of absolute swap contrasts | Unrelated prompt–trace pairs | Two swapped-input forward evaluations | Incoherent inputs and unidentified causal effects |
Appendix figures & tables12 assets
Supplementary material from the paper’s appendix.
Appendix
| Question | Vanilla Model | Model |
|---|---|---|
| A customs officer counts 2 containers of 5 vehicles each on day one. By day two the port total is 30 vehicles. How many containers arrived on day two? | • Total vehicles on the first day = 2 containers × 5 vehicles/container = 10 vehicles. • Total vehicles on the second day = 30 vehicles. • 30 vehicles 10 vehicles = 20 vehicles. • 20 vehicles / 5 vehicles/container = 4 containers. | • Total vehicles on the first day = 2 * 5 = 10. • Total vehicles on the second day = 30 10 = 20. • 20 / 5 = 4. |
| Toula bought 3 dozen donuts at 80/dozen, and 6 dozen mini cheesecakes at $55/dozen. What was the total cost? | • 3 dozen donuts = 3 × 204 • 2 dozen mini cupcakes = 2 × 160 • 6 dozen mini cheesecakes = 6 × 330 • Total cost = 160 + 694 | • 3 dozen donuts = 3 × 68 = 204 • 2 dozen mini cupcakes = 2 × 80 = 160 • 6 dozen mini cheesecakes = 6 × 55 = 330 • 204 + 160 + 330 = 694 |
| Nancy owes 2.00 fee. How much does she pay in total? | • Nancy owes 8 books × 4.00. • Adding the 4.00 + 6.00. | • Total cost of overdue books: 8 × 0.50 = 4.00 • Add the flat fee: 4.00 + 2.00 = 6.00 |
| A factory sold 10 tractors/day at 220 each. What percent more does it make per day now? | • The factory made 10 tractors a day at 100 = 220 each, so they make 5 × 1100. • The difference is -1000 = 100 / $1000) × 100 = 10%. | • Old price per tractor: 220. New number of silos: 5. New total: 2205 = 1100. • New total: 1100 1000 = 100. • 100/1000100 = 10%. |
| Jillian’s handbag cost 80. How much did the bag cost? | • Her shoes cost 80 = -20 = -20 = $220. | • Her shoes cost -$ 20 = 220. |
| Question | Vanilla Model | Model |
|---|---|---|
| A plane travels 1200 miles in 3 hours. At the same rate, how many additional hours to travel an additional 2000 miles? | • The plane travels 1200 miles in 3 hours. • To find the rate, divide the distance by the time: 1200 miles / 3 hours = 400 miles per hour. • To find the time to travel 2000 miles, divide the distance by the rate: 2000 miles / 400 miles per hour = 5 hours. | • Distance = 1200 miles. Rate = 1200/3 = 400 miles per hour. • Additional distance = 2000 miles. Additional time = 2000/400 = 5 hours. |
| John’s cooking class meets 4 times/week for 2 hours, for 6 weeks. He learns a recipe every 1.5 hours. How many recipes does he learn? | • Number of class sessions = 4 sessions/week × 6 weeks = 24 sessions. • Each session = 2 hours. Total class time = 24 sessions × 2 hours/session = 48 hours. • Number of hours learned = 48 hours. Number of recipes learned = 48 hours / 1.5 hours/recipe = 32 recipes. | • Total class time = 4 * 2 * 6 = 48 hours. • 48 / 1.5 = 32 recipes. |
| Mrs. Jameson’s bamboo grows 30 inches/day and is currently 20 feet tall. In how many days will it be 600 inches? | • First, convert the height from feet to inches: 20 feet × 12 inches/foot = 240 inches. • Now, subtract the current height from the target height: 600 inches 240 inches = 360 inches. • Divide the remaining inches by the growth rate: 360 inches / 30 inches/day = 12 days. | • Convert 20 feet to inches: 20 * 12 = 240 inches. • 600 240 = 360 inches. • 360 / 30 = 12. |
| Jon’s car needs a tune-up every 1000 miles. He drives 100 miles/day for a 30-day month. How many tune-ups does he need? | • Jon drives 100 miles a day for 30 days. • 100 miles/day × 30 days = 3000 miles. • Since Jon needs a tune-up every 1000 miles, we divide 3000 miles by 1000 miles to find the number of tune-ups. • 3000 miles 1000 miles/tune-up = 3 tune-ups. | • Total miles driven in a month = 100 * 30 = 3000. • Number of tune-ups = 3000 / 1000 = 3. |
| Sid traveled 110 miles in 2 hours, then an additional 140 miles in 3 hours. What was his average speed? | • First, calculate the total distance traveled: 110 miles + 140 miles = 250 miles. • Then, calculate the total time taken: 2 hours + 3 hours = 5 hours. • Average speed = Total distance / Total time = 250 miles / 5 hours = 50 miles per hour. | • Total distance = 110 + 140 = 250 miles. • Total time = 2 + 3 = 5 hours. • Average speed = 250/5 = 50. |
| Score | Spearman with accuracy |
|---|---|
| Log-probability | |
| Self-consistency ( ) |
| EA-AUC | Shuf. | Mist. | Reph. | |
|---|---|---|---|---|
| Method | Full fwd | Bwd | Time ratio |
|---|---|---|---|
| Vanilla | 0 | 0 | |
| CoT Decomposition | 0 | 0 | |
| 0 | 1 | ||
| 1 | 0 | ||
| 0 | 0 | ||
| 2 | 0 |
| Item | Weight | Description |
|---|---|---|
| Solution wrapper | 0.22 | Generation starts with <solution> and ends with </solution> and nothing outside |
| Single solution | 0.15 | Generation contains exactly one <solution> |
| Thinking once | 0.12 | Exactly one <thinking>...</thinking> inside solution |
| Steps block | 0.15 | Generation contains at least two <step>...</step> inside thinking. Only awarded if no stray text |
| Steps no stray text | 0.18 | Inside <thinking> , ignoring whitespace, there should be nothing outside <step> blocks |
| Single answer | 0.14 | Generation contains exactly one <answer>...</answer> inside solution |